How do you write equations for perpendicular lines?
I will assume you know how to find the equation of a line if you know the slope and a point on that line.
The key thing to remember is:
Two perpendicular lines have slopes that are opposite reciprocals of each other.
e.g if one has slope -8/13 then its perpendicular has slope of 13/8.
Simply use that fact and you should have no problem.
I understand that much...
The problem that I'm having an issue with is this...
Find an equation of the line that contains (1,8) and is perpendicular to y = 3/4x + 1.
I always get the answer y = -4/3 + 8, but the answer is y = -4/3 + 9 1/3.
I just don't understand how they get the "9 1/3" part.
Ok
the slope of the old line is 3/4 so the slope of the new line must be -4/3
let it be y = (-4/3)x + b
but the point (1,8) is on it, so
8 = -4/3(1)+b
8+4/3=b
b=28/3
therefore y = -4/3 x + 28/3 or
y = -4/3 x + 9 1/3
You said "I always get the answer y = -4/3 + 8, but the answer is y = -4/3 + 9 1/3."
Your equation makes no sense, it has no x, and why did you put the 8 where the y-intercept should be????
I forgot to put the x's in, sorry. I MENT to say "y = (-4/3)x + 8" and "y = (-4/3) + 9 1-3"
I put the 8 where the y-intercept should go because that's where it said to put it in the packet that I have for over the summer Geometry homework. It showed step by step instructions and that's where it said to put it. Maybe it's teachng it wrong then.
Thanks, the way you did it made WAYY more sence than the way the page explained it. They made it way more complicated than it truley is. I'm definatly going to come here for any other homework help.
I am pretty sure that your packet says to put Y-intercept where you put the 8
But the 8 was NOT the y-intercept, the point was (1,8) so it could not have been on the y axis.
You should recognize the y-intercept if the point has the form (0,?)
If you are given any point, simply replace the x and y of your equation with the x and y values of the given point.
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